What is the integral of (4x)/(sqrt(-x^2-8x-7)) ?

The integral of (4x)/(sqrt(-x^2-8x-7)) is 4(-sqrt(-x^2-8x-7)-4arcsin(1/3 (x+4)))+C

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integral of (4x)/(sqrt(-x^2-8x-7))

\int \frac{4 x}{- x ^{2} - 8 x - 7} d x

4 (- - x^{2} - 8 x - 7 - 4 arcsin (\frac{1}{3} (x + 4))) + C

Solution steps

\int \frac{4 x}{- x ^{2} - 8 x - 7} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{x}{- x ^{2} - 8 x - 7} d x

Complete the square

- x^{2} - 8 x - 7 : - (x + 4)^{2} + 9

= 4 \cdot \int \frac{x}{- ( x + 4 ) ^{2} + 9} d x

Apply u-substitution

= 4 \cdot \int \frac{u - 4}{- u ^{2} + 9} d u

Expand

\frac{u - 4}{- u ^{2} + 9} : \frac{u}{- u ^{2} + 9} - \frac{4}{- u ^{2} + 9}

= 4 \cdot \int \frac{u}{- u ^{2} + 9} - \frac{4}{- u ^{2} + 9} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int \frac{u}{- u ^{2} + 9} d u - \int \frac{4}{- u ^{2} + 9} d u)

\int \frac{u}{- u ^{2} + 9} d u = - - u^{2} + 9

\int \frac{4}{- u ^{2} + 9} d u = 4 arcsin (\frac{1}{3} u)

= 4 (- - u^{2} + 9 - 4 arcsin (\frac{1}{3} u))

Substitute back

u = x + 4

= 4 (- - (x + 4)^{2} + 9 - 4 arcsin (\frac{1}{3} (x + 4)))

Expand

- (x + 4)^{2} + 9 : - x^{2} - 8 x - 7

= 4 (- - x^{2} - 8 x - 7 - 4 arcsin (\frac{1}{3} (x + 4)))

Add a constant to the solution

= 4 (- - x^{2} - 8 x - 7 - 4 arcsin (\frac{1}{3} (x + 4))) + C

What is the integral of (4x)/(sqrt(-x^2-8x-7)) ?
The integral of (4x)/(sqrt(-x^2-8x-7)) is 4(-sqrt(-x^2-8x-7)-4arcsin(1/3 (x+4)))+C